An algebra problem by Aly Ahmed

Algebra Level pending

If a > 2 b > 0 a>2b>0 , what is the minimal value of a 2 + 1 2 a b + 1 a ( a − 2 b ) ? a^2 + \dfrac1{2ab} + \dfrac1{a(a-2b)} ?


The answer is 4.

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

Advait Nene
Jun 7, 2018

Disclaimer: I haven't formally learned multivariable calculus, so there may be some technicalities I miss in this section.

Let's define a function f ( a , b ) f(a,b) :

f ( a , b ) = a 2 + 1 2 a b + 1 a ( a − 2 b ) f(a,b)=a^{2}+\frac{1}{2ab}+\frac{1}{a(a-2b)}

We will now take the partial derivatives of f f with respect to a a and b b , set both equal to 0 0 and solve the system of equations.

∂ f ∂ a = 2 a − 1 2 a 2 b − 2 a − 2 b ( a 2 − 2 a b ) 2 ∂ f ∂ b = − 1 2 a b 2 + 2 a ( a 2 − 2 a b ) 2 \begin{aligned} \frac{\partial f}{\partial a}&=2a-\frac{1}{2a^{2}b}-\frac{2a-2b}{(a^{2}-2ab)^{2}}\\ \frac{\partial f}{\partial b}&=-\frac{1}{2ab^{2}}+\frac{2a}{(a^{2}-2ab)^{2}}\\ \end{aligned}

{ 2 a − 1 2 a 2 b − 2 a − 2 b ( a 2 − 2 a b ) 2 = 0 − 1 2 a b 2 + 2 ( a 2 − 2 a b ) 2 a = 0 \begin{cases} 2a-\frac{1}{2a^{2}b}-\frac{2a-2b}{(a^{2}-2ab)^{2}}=0\\ -\frac{1}{2ab^{2}}+\frac{2}{(a^{2}-2ab)^{2a}}=0\\ \end{cases}

After solving, we get that a = 2 a=\sqrt{2} and b = 2 4 b=\frac{\sqrt{2}}{4} . Then, f ( 2 , 2 4 ) = 4 f(\sqrt{2},\frac{\sqrt{2}}{4})=4

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